Reversable process with adiabatic expansion

[osheari1]

Homework Statement

One mole of a monatomic ideal gas is taken through the reversible cycle shown below.
http://img508.imageshack.us/img508/2595/newbitmapimageqm.png
process bc is an adiabatic expansion, with P_b = 10.00 atm and V_b = 10^-3 m^3.
Find (a) the energy added to the gas as heat. (b) find the energy leaving the gas as heat, (c) the net work done by the gas, and (d) the efficiency of the cycle.

Homework Equations

PV=nRT (ideal gas)
PV^(gamma) = constant
gamma = C_p/C_v
V^(gamma-1)T = constant
U = (3/2)nRT (for a monatomic gas)

The Attempt at a Solution

I really don't know how to start this problem
I know for the bc section of the cycle Q = 0 because its an adiabatic expansion.
I also solved for the temperature at point b T = PV/(nR) = (1.0132*10^6Pa)(10^-3m^3)/(1mole)(8.315) = 121.85 K

from here I am not sure what to do because I am not given gamma or the pressure at a or c


Your help is much appreciated

 

[Andrew Mason]

Homework Statement

One mole of a monatomic ideal gas is taken through the reversible cycle shown below.
http://img508.imageshack.us/img508/2595/newbitmapimageqm.png
process bc is an adiabatic expansion, with P_b = 10.00 atm and V_b = 10^-3 m^3.
Find (a) the energy added to the gas as heat. (b) find the energy leaving the gas as heat, (c) the net work done by the gas, and (d) the efficiency of the cycle.

Homework Equations

PV=nRT (ideal gas)
PV^(gamma) = constant
gamma = C_p/C_v
V^(gamma-1)T = constant
U = (3/2)nRT (for a monatomic gas)

The Attempt at a Solution

I really don't know how to start this problem
I know for the bc section of the cycle Q = 0 because its an adiabatic expansion.
I also solved for the temperature at point b T = PV/(nR) = (1.0132*10^6Pa)(10^-3m^3)/(1mole)(8.315) = 121.85 K

from here I am not sure what to do because I am not given gamma or the pressure at a or c

[Note: A reversible cycle has to be in equilibrium with its surroundings at all time. This means heat flow can only occur when there is an infinitessimal temperature difference between the system and the hot or cold reservoirs. So heatflow has to occur isothermally. This cannot occur on the isochoric or isobaric parts of this cycle so I fail to see how this can be a reversible cycle. But that is just an aside.]

You have to determine the temperatures at a, b and c.

First, determine the temperature at b using the ideal gas law.

Second, determine the temperature and pressure at c using the adiabatic condition.

Then determine the temperature at a from that.

Since a-b is isochoric you can work out the heatflow into the gas from the change in temperature (using what specific heat capacity and what value?)

Since c-a is isobaric, you can work out the heatflow out of the gas from the change in temperature (using what specific heat capacity and what value?).

Since b-c is adiabatic, what can you say about the heat flow from b-c?

AM

 

[osheari1]

after working it out I ended up with temperature values of T_a = 3.86K, T_b = 121.85, T_c = 30.88. however, these values seems extremely low, but I just assumed it was because of the very small volume.
I determined these values by calculated gamma. sense the gas is monatomic and ideal, 5/2R = C_p and C_p = C_v +R so Gamma = 5/2/(5/2-1) = 1.66

so, the heat flowing into the system is from the Isochoric process, Q = U + 0 = 3/2nRT = 1471.6J
the heat flowing out of the system is from the isobaric process, Q = 5/2nRT = -561.68J

The only work done by the process is from the adiabatic expansion and the isobaric process, so Wnet = Wbc + Wca = -(Ubc) + (Qca - Uca) = -(3/2*8.315(30.88-121.85)) + (-561.67 - 3/2nRT) = 1134.62 +(337 - 561.67) = 909.95 J

 

[Andrew Mason]

after working it out I ended up with temperature values of T_a = 3.86K, T_b = 121.85, T_c = 30.88. however, these values seems extremely low, but I just assumed it was because of the very small volume.
I determined these values by calculated gamma. sense the gas is monatomic and ideal, 5/2R = C_p and C_p = C_v +R so Gamma = 5/2/(5/2-1) = 1.66

so, the heat flowing into the system is from the Isochoric process, Q = U + 0 = 3/2nRT = 1471.6J
the heat flowing out of the system is from the isobaric process, Q = 5/2nRT = -561.68J

The only work done by the process is from the adiabatic expansion and the isobaric process, so Wnet = Wbc + Wca = -(Ubc) + (Qca - Uca) = -(3/2*8.315(30.88-121.85)) + (-561.67 - 3/2nRT) = 1134.62 +(337 - 561.67) = 909.95 J

That looks right. You don't have to do so much calculation to find the work. By the first law, since there is no change in U in one cycle, so the work done by the gas must equal the net heat flow: $W = \Delta Q = Q_{in} - Q_{out} = 1471.6-561.7 = 909.9 J$

AM